All categories

3 years ago

Can someone help pls omg

Mathematics

1 answer:

hodyreva [135]3 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

Sum the sides of both polygons and equate.

(20)

x + 4 + x + 2 + x + 5 = 2(x + 1) + 2(x + 3) ← distribute parenthesis

3x + 11 = 2x + 2 + 2x + 6

3x + 11 = 4x + 8 ( subtract 3x from both sides )

11 = x + 8 ( subtract 8 from both sides )

3 = x

(21)

5(12x) = x + 7 + 6x + 9 + x + 10

60x = 8x + 26 ( subtract 8x from both sides )

52x = 26 ( divide both sides by 52 )

x = $\frac{26}{52}$ = $\frac{1}{2}$

You might be interested in

Find the arc length of the particle circle

klasskru [66]

Answer:

Length of arc is 2π

Step-by-step explanation:

Length of arc=¤/360 x 2πr

Where:

¤=angle

r=radius

length of arc=90/360 x 2xπx4

Length of arc=1/4 x 8π

Length of arc=8π ➗ 4

Length of arc=2π

4 0

3 years ago

Read 2 more answers

Is this a function? Please help me! I literally cant do this!

julsineya [31]

Answer:

Yes, this is a function.

Step-by-step explanation:

This is a function because every input goes to exactly one output. Although 1 is listed twice in this example, it still goes to the same output of 15.

5 0

3 years ago

At the burger place, 3 mini hamburgers fit in each small to-go bag. if the baldwin family ordered 41 mini hamburgers for the tri

arsen [322]

How man bags were used?

41 / 3 = 13.7

We can say about 14 bags by rounding to the nearest whole number. Therefore, 14 bags were used.

5 0

3 years ago

Read 2 more answers

Consider x+4=x+8 . What values could be substituted for x to make this a true number sentence?

defon

Answer:

There is no value of x that will make this true

Step-by-step explanation:

x+4=x+8

Subtract x from each side

x-x+4=x-x+8

4 = 8

This is never true so there is no solution

8 0

3 years ago

The intensity of a light source at a distance is directly proportional to the strength of the source and inversely proportional

jolli1 [7]

Answer:

x= $\frac{16}{\sqrt[3]{2}+1 }$

Step-by-step explanation:

Q= illumination

I = intensity

Q= I/d^2

Q_total = $\frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}$

= $\frac{I}{x^2}+\frac{2I}{(16-x)^2}$

now Q' = 0

⇒I ${-\frac{2}{x^3}}+\frac{4}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

[/tex] $\frac{1}{x^3} = \frac{2}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

is the required point

5 0

3 years ago